1 Mark Question

Question 11 Mark

In right angled $\triangle \mathrm{ACB}$, If $\angle \mathrm{C}=90^{\circ}, \mathrm{AC}=3, \mathrm{BC}=4$.Find the ratios $\sin A, \sin B, \cos A, \tan B$

Answer

In right angled $\triangle \mathrm{ACB}$, using Pythagoras' theorem,
$\mathrm{AB}^2=\mathrm{AC}^2+\mathrm{BC}^2$
$=3^2+4^2=5^2$
$\therefore \mathrm{AB}=5$
$\sin \mathrm{A}=\frac{B C}{A B}=\frac{4}{5}$
$and \sin \mathrm{B}=\frac{A C}{A B}=\frac{3}{5}$
$\cos \mathrm{A}=\frac{A C}{A B}=\frac{3}{5}$
$\tan \mathrm{B}=\frac{A C}{B C}=\frac{3}{4}$

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Question 21 Mark

Find the value of $\frac{\cos 56^{\circ}}{\sin 34^{\circ}}$

Answer

$56^{\circ}+34^{\circ}=90^{\circ}$ means 56 and 34 are the measures of complimentary angles.
$
\begin{aligned}
\sin \theta & =\cos (90-\theta) \\
\therefore \quad \sin 34^{\circ} & =\cos (90-34)^{\circ}=\cos 56^{\circ} \\
\therefore \quad & \frac{\cos 56^{\circ}}{\sin 34^{\circ}}=\frac{\cos 56^{\circ}}{\cos 56^{\circ}}=1
\end{aligned}
$

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Question 31 Mark

Find the value of 2tan 45° + cos 30° - sin 60°

Answer

$2 \tan 45^{\circ}+\cos 30^{\circ}-\sin 60^{\circ}$
$=2 \times 1+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$
$=2+0$
$=2$

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Maths 1 Mark Question

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